Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Today I proved that if $V$ is a normed space with norm $\|\cdot\|$ then I can define a norm on $V \times V$ that induces the same topology as the product topology as follows: $\| (v,w) \|_{V \times V} = \|v\| + \|w\|$.

I think I can do the same for an infinite product $V^{\mathbb N}$ by defining $\|(v_n)\|_{\mathbb N} = \sum_{n=0}^\infty \frac{1}{2^n} \|v_n\|$ and I proved it using the proof of the case $V \times V$ and changing some minor things.

Can you confirm that this result is correct? Thanks.

share|cite|improve this question
The series may be divergent. – Davide Giraudo Jul 7 '12 at 21:09
You need to have some restrictions on $V^{\mathbb N}$, such as requiring $\sum_{n=0}^\infty \|v_n\| < \infty$. Just as you would with $\mathbb{R}^{\mathbb N}$ (ie, the $l_p$ spaces). – copper.hat Jul 7 '12 at 21:14
I'm not sure you can give a norm for the infinite product topology, as it satisfies the Heine-Borel property (bounded, closed sets are compact) but is infinite-dimensional. – Cocopuffs Jul 7 '12 at 21:23
up vote 5 down vote accepted

Proposition. If $X_i, i \in \mathbb{N}$ is any sequence of nonzero topological vector spaces, then the product topology on $X = \prod_i X_i$ is not normable.

Proof. Suppose to the contrary there is a norm $\|\cdot\|$ on $X$ which induces the product topology. Let $B \subset X$ be the open unit ball of $\|\cdot\|$. $B$ is open in the product topology and contains 0, so we can find a basic open neighborhood of 0 which is contained in $B$. That is, there is a set of the form $U = U_1 \times U_2 \times \dots \times U_n \times X_{n+1} \times \cdots$, where $U_i \subset X_i$ is open and nonempty, such that $0 \in U \subset B$. Choose any nonzero $x_{n+1} \in X_{n+1}$ and set $x = (0,\dots, 0, x_{n+1}, 0, \dots)$. Then for any $t \in \mathbb{R}$ we have $tx \in U \subset B$. If we take $t = 2/\|x\|$, we have $\|tx\| = 2$ which is absurd since $B$ is the unit ball.

In short, the problem is that every open set of the product topology contains a line, and balls of a norm do not.

share|cite|improve this answer

The correct question is not "can I put a norm on the ordinary product?" but "what should the product of infinitely many Banach spaces be?" and then one asks what its relation is to the ordinary product. One answer is to take the subspace of $V^{\mathbb{N}}$ consisting of all sequences $v_i$ such that $$\sum_{i=1}^{\infty} \| v_i \|$$

converges, with the above as norm. This is the coproduct of countably many copies of $V$ in the category $\text{Ban}_1$ of Banach spaces and weak contractions. Another is to take the subspace of $V^{\mathbb{N}}$ consisting of all sequences $v_i$ such that $$\sup_i \| v_i \|$$

exists, with the above as norm. This is the product of countably many copies of $V$ in $\text{Ban}_1$.

It follows that the forgetful functor $\text{Ban}_1 \to \text{Set}$ sending a Banach space to its set of vectors preserves neither products nor coproducts. It is better to use another forgetful functor $\text{Hom}(1, -)$ sending a Banach space to its unit ball, which preserves products (in fact since it is representable it preserves all limits).

share|cite|improve this answer

The topology on $V^{\mathbb{N}}$ is generated by the Fréchet metric $$d((x_j),(y_j)) = \sum_{k=0}^{\infty} \frac{|x_k - y_k|}{2^k(1 + |x_k - y_k|)}$$ which seems to be in the spirit of what you suggested. This isn't induced by any norm, though.

share|cite|improve this answer

Suppose $\|\cdot \|$ is a norm on $V^{\mathbb N}$ and induces the product topology. Let $p_i$ denote projection onto the $i^{th}$ coordinate. Since projections are continuous, for any $y\in V^{\mathbb N}$ we have that the function $$\|\cdot \|_{y,j}: \{x\in V^{\mathbb N}:p_i(x)=p_i(y),\forall i\neq j\}\to V$$ defined by $\|x\|_{y,j}=\|x\|$ is a norm on $\{x\in V^{\mathbb N}:p_i(x)=p_i(y),\forall i\neq j\}\cong V$. Choose some $x_1\in V$ such that $\|(x_1,0,\ldots)\|\geq 1$. Once you've chosen $x_1,\ldots,x_{n-1}$ let $y=(x_1,\ldots,x_{n-1},0,\ldots)$, which by hypothesis satisfies $\|y\|\geq n-1$. Choose some $x\in \{x\in V^{\mathbb N}:p_i(x)=p_i(y),\forall i\neq j\}$ such that $\|x\|\geq n$, and let $x_n=p_n(x)$. The sequence $(x_1,x_2,\ldots)$ constructed in this manner must have infinite norm, which is impossible. Thus $V^{\mathbb N}$ has no norm.

Edit: As Nate pointed out in the comments, my answer was incorrect. I have rewritten it accordingly.

share|cite|improve this answer
Indeed, any countable product of metric spaces $(X_i, d_i)$ is metrizable; the standard trick is to use the metric $$d(x,y) = \sum_{i=1}^\infty \frac{d_i(x_i, y_i)}{2^i(1+d_i(x_i,y_i))}.$$ I'm not sure what you mean by "the proof above works just as well". – Nate Eldredge Jul 7 '12 at 22:41
I think your first sentence is erroneous: $U^{\mathbb{N}}$ is not open in the product topology (unless $U=V$). – Nate Eldredge Jul 7 '12 at 22:44
@NateEldredge You're quite right. Sometimes I forget the basis for the infinite product topology. Editing. – Alex Becker Jul 7 '12 at 22:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.